Question

Solve ${\tan }^2\theta -2\sin \theta =0$

Answer 1

Given, ${\tan }^2\theta -2\sin \theta =0\Rightarrow {\sin }^2\theta -2{\cos }^2\theta \sin \theta =0\Rightarrow {\sin }^2\theta -2(1-{\sin }^2\theta )\sin \theta =0\Rightarrow {\sin }^2\theta -2\sin \theta +2{\sin }^3\theta =0\Rightarrow \sin \theta \left(2{\sin }^2\theta +\sin \theta -2\right)=0$
$\Rightarrow \sin \theta =0$ or $2{\sin }^2\theta +\sin \theta -2=0$
$\Rightarrow \theta =n\pi$ or $\sin \theta =\frac{-1\pm \sqrt{1+16}}{2}$
$\Rightarrow \theta =n\pi$ or $\sin \theta =\frac{1\pm \sqrt{17}}{2}$
As $-1\le \sin x\le 1$
$\therefore \theta =n\pi$